If the sum of n consecutive integers is 0, which of the following must

Let the consecutive no. as mentioned below:
-2,-1,0,1 and 2 - sum is zero.
statement I: N cannot be even. N is always odd.In above N=5.
Statement II: N is always odd.
STATEMENT III: average above is Zero.
Correct answer is both II and III.

Corrct Option is E

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Solution

Given:

• Sum of n consecutive integers = 0

To find:

• Which of the given options is always true?

Approach and Working:

• Let the ‘n’ consecutive terms be {a, a+1, a+2, a+3, …., a+n-1}, where a is an integer.
• Sum of all these n consecutive terms = \(a*n + (1+2+3+ … + n-1) = an + (n-1) * \frac{(n-1+1)}{2} = an + \frac{n(n-1)}{2}\)

o Given this sum = 0
o That is, \(an + \frac{n(n-1)}{2} = 0\)
o Implies, \(a + \frac{(n-1)}{2} = 0\)
o Which gives, n = -2a +1 …………... (1)

• Thus, n = 2p + 1, where p = -a = integer (since, a is an integer)

Therefore, n is an odd number

• Average of n consecutive integers = \(\frac{(first term + last term)}{2} = \frac{(a + a+n-1)}{2}\)
• Substituting the value of n from equation (1), we get,

o Average of n terms = \(\frac{[2a + (-2a +1) -1]}{2} = 0\)

Therefore, statements II and III must be true.

Hence, the correct answer is option E.

Answer: E

In this question, I took the following consecutive integers -

1,2,3,4,5

Option 1 - is wrong as the total is 15 ()

Option 2- 15 is an Odd Number

Option 3 - I am getting 1+5/2 = 3

and not zero

Can anyone tell where am going wrong?

Hi, VikramMehta

In order to test the numbers, you need to satisfy the constraint of the sum of 0. It says the sum of n consecutive integers is 0

Therefore you need to have 0 in the middle and the same numbers in negative and positive so they cancel each other
For example :

-1 0 +1 => the sum of the 3 consecutive integers is 0

-3 -2 -1 0 +1 +2 +3 ==> the sum of the 7 consecutive integers is 0

As you can see , the set will always be odd as you have 0 in the middle standing and the "negative/positive twins" on each opposite side, it's symmetric.

Now, because the sum is always 0, whether you have 7 consecutive integers or 9, or even 121 consecutive integers, the average will always be 0 because 0/7=0 just like 0/121=0

Let me know if you need further explanation!

Why cant n=0? Sum of 0 numbers is always zero. Would make 2 false

What if n is zero, zero consecutive numbers, then the sequence will have no numbers and n would be neither odd nor even

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Not a valid case as the question clearly requires the presence of a certain number of integers. Also, zero is even

To determine which statements are true, let's consider the sum of n consecutive integers equal to 0.

Let's assume the first integer in the sequence is x. Then the sum of n consecutive integers can be expressed as:

x + (x+1) + (x+2) + ... + (x+n-1) = 0

We can simplify this equation by rearranging and combining like terms:

nx + 1 + 2 + ... + (n-1) = 0
nx + (1 + 2 + ... + (n-1)) = 0
nx + ((n-1)(n-1+1))/2 = 0
nx + ((n-1)(n))/2 = 0
nx + (n^2 - n)/2 = 0

Now, let's analyze the given statements:

I. n is an even number:
If n is an even number, then we can express it as n = 2k, where k is an integer. Substituting this into the equation above:

x(2k) + ((2k)^2 - 2k)/2 = 0
2kx + (4k^2 - 2k)/2 = 0
2kx + 2k^2 - k = 0
2k(x + k) = k

The left-hand side (LHS) is even, but the right-hand side (RHS) is odd (k is not divisible by 2). Therefore, statement I cannot be true.

II. n is an odd number:
If n is an odd number, then we can express it as n = 2k + 1, where k is an integer. Substituting this into the equation above:

x(2k + 1) + ((2k + 1)^2 - (2k + 1))/2 = 0
2kx + x + (4k^2 + 4k + 1 - 2k - 1)/2 = 0
2kx + x + (4k^2 + 2k)/2 = 0
2kx + x + 2k^2 + k = 0
2k(x + 1) + (k + 2k^2) = 0
2k(x + 1) + k(1 + 2k) = 0

Both the LHS and RHS are even since they involve the multiplication of an even number (2k) with an integer. Therefore, statement II can be true.

III. The average (arithmetic mean) of the n integers is 0:
To find the average of n consecutive integers, we divide the sum by n:

(x + (x+1) + (x+2) + ... + (x+n-1))/n = 0

If we simplify this equation:

nx + (1 + 2 + ... + (n-1)) = 0
nx + ((n-1)(n))/2 = 0

This equation is the same as the one we obtained earlier. Therefore, the average of the n integers will be 0 if the sum is 0. So, statement III can be true.

Based on our analysis, the correct answer is:

E. II and III

Why including 0 is mandatory? The sum can be 0 in even case as well [1,-1,2,-2]